A region in $\mathbb{R}^n$ is a subset $\Omega$ of $\mathbb{R}^n$ for which $\exists$ a function $f:\mathbb{R}^n\ \rightarrow\ \mathbb{R}$ with following properties:
(i) all partial derivatives of f are continuous,
(ii) $\Omega = \left \{\underline{x} \in \mathbb{R}^n \mid \ f\left(\underline{x}\right)<0 \right \}$ and
(iii)$\nabla f\left(\underline{x}\right) \neq 0\, \forall \underline{x} \in \partial{\Omega}$ where $\partial{\Omega} := \left \{\underline{x} \in \mathbb{R}^n \mid \ f\left(\underline{x}\right)=0 \right \}$ is the boundary of $\Omega$
$f$ is called a defining function of the set $\Omega$.
Condition (iii) in the definition of a region allows us to define the outward unit normal $\underline{n}$ to $\Omega$ by $\underline{n}\left(\underline{p}\right) := \frac{\nabla f\left(\underline{p}\right)}{ \left\|{\nabla f\left(\underline{p}\right)}\right\|} \forall \underline{p} \in \partial{\Omega}$
Observe that, if $\epsilon>0$ is sufficiently small then $f(\underline{p}+\epsilon\underline{n}\left(\underline{p}\right))>0$. (How do I prove this?) and therefore $\underline{p}+\epsilon\underline{n}\left(\underline{p}\right) \notin \Omega$ (this makes sense)
You may consider $$ \lim_{\epsilon\to 0}\frac{f(p+\epsilon n(p))-f(p)}{\epsilon}=\langle \nabla f(p),n(p)\rangle =\frac{\langle \nabla f(p),\nabla f(p) \rangle}{\|\nabla f(p) \|} = \|\nabla f(p) \| > 0. $$ So for $\epsilon$ small, you have $$ f(p+\epsilon(p))>f(p)=0. $$